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Calculus1 10 Online
OpenStudy (anonymous):

The density of material at any point in a solid cylinder, y^2 + z^2 ≤ a^2, 0 ≤ x ≤ b is proportional to the distance of the point from the x-axis. Find the density. Use k as the proportionality constant. δ(y, z) = δ(r) =

OpenStudy (anonymous):

I think I got the density

OpenStudy (anonymous):

\[k \sqrt{y^2+z^2}\]

OpenStudy (anonymous):

what do you think @wio ?

OpenStudy (anonymous):

It would help if you draw it.

OpenStudy (anonymous):

But I believe that your density function is correct

OpenStudy (anonymous):

how would I change it to polar coordinates? @wio

OpenStudy (anonymous):

\[ r^2 = y^2+z^2 \]

OpenStudy (anonymous):

\[ y=r\sin\theta,\, z=r\cos\theta \]

OpenStudy (anonymous):

And \(x\) remains the same as the height component

OpenStudy (anonymous):

Since it is a cylinder \(0\leq \theta\lt 2\pi\). Since \(x\) acts as the height, it can remain the same.

OpenStudy (anonymous):

thanks @wio :)

OpenStudy (anonymous):

wait so I don't haven an x on my density funtion @wio

OpenStudy (anonymous):

\[k \sqrt{rsin \theta+rcos \theta} \]

OpenStudy (anonymous):

No, you don't, because the \(x\) value doesn't change the density

OpenStudy (anonymous):

Changing the \(x\) value doesn't change the distance from the \(x\) axis

OpenStudy (anonymous):

\[ \delta (y,z) = k\sqrt{y^2+z^2} = k\sqrt{r^2} = kr \]

OpenStudy (anonymous):

Well \(k\sqrt{r^2} = k|r| = kr\), we can say \(|r|=r\) because by design \(r\geq0\).

OpenStudy (anonymous):

I don't like how they have \(\delta (y,z)\) and \(\delta (r)\). They are two different functions. I suppose that is me nitpicking though.

OpenStudy (anonymous):

For example \(\delta(r,r) = k\sqrt{r^2+r^2} = kr\sqrt 2\neq kr=\delta(r)\).

OpenStudy (anonymous):

for the second part of this question they ask to set up an integral

OpenStudy (anonymous):

\[\int\limits\limits_{?}^{?}\int\limits\limits_{?}^{?}\int\limits\limits_{?}^{?}f(x,r,\theta)dx dr d \theta \]

OpenStudy (anonymous):

so it would be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{6}\int\limits_{0}^{5}kr dx d r d \theta \]

OpenStudy (anonymous):

cylinder is radius 6 and length 5 @wio

OpenStudy (anonymous):

Remember the Jacobian

OpenStudy (anonymous):

\[ dydz =rdrd\theta \]

OpenStudy (anonymous):

what's the jacobian?

OpenStudy (anonymous):

It's the scaling factor of the differentials

OpenStudy (anonymous):

so it would be \[dx r dr d \theta \]

OpenStudy (anonymous):

what whould the k value be?

OpenStudy (anonymous):

proportinality constant

OpenStudy (anonymous):

they ask for exact answer by having k I can not find an exact answer

OpenStudy (anonymous):

You can get an exact answer in terms of \(k\)

OpenStudy (anonymous):

can k just be x

OpenStudy (anonymous):

No, because that wouldn't work

OpenStudy (anonymous):

\(k\) is constant with respect to \(x,y,z\)

OpenStudy (anonymous):

i'm so confuse I'm sorry

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